Question

Let $N$ denote the set of all natural numbers and $R$ be the relation on $N\times N$ defined by $(a,b)R(c,d),$ if $ad(b+c)=bc(a+d),$ then show that $R$ is an equivalence relation.

Answer 1

$(a,b)R(c,d)\iff ad(b+c)=bc(a+d)$

$ab(b+a)=ba(a+b)\Rightarrow (a,b)R(a,b)$

Thus, the given relation is reflexive

$(a,b)R(c,d)\Rightarrow ad(b+c)=bc(a+d)ad(b+c)=bc(a+d)\Rightarrow cb(d+a)=da(c+b)\Rightarrow (c,d)R(a,b)\therefore (a,b)R(c,d)\Rightarrow (c,d)R(a,b)$

Thus, the given relation is symmetric

$(a,b)R(c,d)\iff ad(b+c)=bc(a+d)\Rightarrow \frac{1}{b}+\frac{1}{c}=\frac{1}{a}+\frac{1}{d}.....\left(1\right)(c,d)R(e,f)\iff cf(d+e)=de(c+f)\Rightarrow \frac{1}{f}+\frac{1}{c}=\frac{1}{e}+\frac{1}{d}....\left(2\right)\left(2\right)-\left(1\right)\Rightarrow \frac{1}{f}-\frac{1}{b}=\frac{1}{e}-\frac{1}{a}\Rightarrow \frac{1}{f}+\frac{1}{a}=\frac{1}{e}+\frac{1}{b}\Rightarrow af(b+e)=be(a+f)\Rightarrow (a,b)R(e,f)\therefore (a,b)R(c,d)\&(c,d)R(e,f)\Rightarrow (a,b)R(e,f)$

Thus, the given relation is transitive.

Thus, the given relation is an equivalence relation.